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2 years ago

Indicate the charge the following elements as they achieve the noble gas configuration.

Chemistry

1 answer:

AveGali [126]2 years ago

3 0

Answer:

See explanation

Explanation:

Ga is in group 13 hence it must loose three electrons to form Ga^3+ in order to achieve the noble gas configuration because it has three electrons on its outermost shell.

O is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form O^2- since oxygen has six electrons on its outermost shell.

Br in group 17 has seven electrons in its outermost shell hence it must form Br^- (gain one electron) in order to attain the noble gas configuration.

P in group 15 must accept three electrons and form P^3- in order to attain the noble gas configuration since it has five electrons on its outermost shell.

S is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form S^2- since sulphur has six electrons on its outermost shell.

Mg in group 2 has two electrons on its outermost shell and must loose both to attain the noble gas configuration forming Mg^2+.

Al is in group 13 hence it must loose three electrons to form Al^3+ in order to achieve the noble gas configuration because it has three electrons on its outermost shell.

Se is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form Se^2- since selenium has six electrons on its outermost shell.

Lithium is in group 1 and must loose its only outermost electron in order to attain the noble gas configuration to form Li^+.

Rb is in group 1 and must loose its only outermost electron in order to attain the noble gas configuration to form Rb^+.

As in group 15 must accept three electrons and form As^3- in order to attain the noble gas configuration since it has five electrons on its outermost shell.

I in group 17 has seven electrons in its outermost shell hence it must form I^- (gain one electron) in order to attain the noble gas configuration.

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Based on the balanced equation 2C2H2 + 5O2 → 4CO2 + 2H2O calculate the number of excess reagent units remaining when 52 C2H2 mol

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Answer:

20 molecules of oxygen gas remains after the reaction.

Explanation:

$2C_2H_2 + 5O_2\rightarrow 4CO_2 + 2H_2O$

Molecules of ethyne = 52

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According to reaction, 2 molecules of ethyne reacts with 5 molecules of oxygen gas.

Then 52 molecules of ethyne will react with:

$\frac{5}{2}\times 52=130 molecules$ of oxygen gas.

As we can see that we have 150 molecules of oxygen gas, but 52 molecules of ethyne will react with 130 molecules of oxygen gas. So, this means that ethyne is a limiting reagent and oxygen gas is an excessive reagent.

Remaining molecules of recessive reagent = 150 - 130 = 20

20 molecules of oxygen gas remains after the reaction.

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Answer:

Correct answers: 2 and 3

Explanation:

1- correct would be: Isolation of ibuprofen is not dangerous, but it is necessary because only one enantiomer has effect on interaction with biologic diana

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Properties of matter with three example each

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Answer:

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Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which

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Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride = 0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be: $= \dfrac{5 \ kg }{1590 \ kg/m^3}$

= 0.0031 m³

Suppose $\beta$ is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

$\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT$

$In ( \dfrac{V_2}{V_1}) = \beta (T_2-T_1)$

$V_2 = V_1 \times exp (\beta (T_2-T_1))$

$V_2 = 0.0031 \ m^3 \times exp (1.2 \times 10^{-3} \times 20)$

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However; the workdone = -PdV

$W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)$

W = - 7.6 J

The heat energy Q = Δ h

$Q = mC_p(T_2-T_1)$

$Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20$

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

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